§4.prop.3περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι.
About the given circle to circumscribe a triangle equiangular to the given triangle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον τὸ ΔΕΖ· δεῖ δὴ περὶ τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι.
Let ABC be the given circle, and DEF the given triangle; it is required then about the circle ABC to circumscribe a triangle equiangular to the triangle DEF.
Ἐκβεβλήσθω ἡ ΕΖ ἐφʼ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ σημεῖα, καὶ εἰλήφθω τοῦ ΑΒΓ κύκλου κέντρον τὸ Κ, καὶ διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ ΚΒ, καὶ συνεστάτω πρὸς τῇ ΚΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Κ τῇ μὲν ὑπὸ ΔΕΗ γωνίᾳ ἴση ἡ ὑπὸ ΒΚΑ, τῇ δὲ ὑπὸ ΔΖΘ ἴση ἡ ὑπὸ ΒΚΓ, καὶ διὰ τῶν Α, Β, Γ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓ κύκλου αἱ ΛΑΜ, ΜΒΝ, ΝΓΛ.
καὶ ἐπεὶ ἐφάπτονται τοῦ ΑΒΓ κύκλου αἱ ΛΜ, ΜΝ, ΝΛ κατὰ τὰ Α, Β, Γ σημεῖα, ἀπὸ δὲ τοῦ Κ κέντρου ἐπὶ τὰ Α, Β, Γ σημεῖα ἐπεζευγμέναι εἰσὶν αἱ ΚΑ, ΚΒ, ΚΓ, ὀρθαὶ ἄρα εἰσὶν αἱ πρὸς τοῖς Α, Β, Γ σημείοις γωνίαι.
Let EF be produced in both directions to the points G, H, and let the center K of the circle ABC be taken, and let a straight line KB be drawn at random, and let there be constructed on the straight line KB and at the point K on it the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH, and through the points A, B, C let straight lines LAM, MBN, NCL be drawn touching the circle ABC. And since LM, MN, NL touch the circle ABC at the points A, B, C, and from the center K to the points A, B, C the straight lines KA, KB, KC have been joined, therefore the angles at the points A, B, C are right.
καὶ ἐπεὶ τοῦ ΑΜΒΚ τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, ἐπειδήπερ καὶ εἰς δύο τρίγωνα διαιρεῖται τὸ ΑΜΒΚ, καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΚΑΜ, ΚΒΜ γωνίαι, λοιπαὶ ἄρα αἱ ὑπὸ ΑΚΒ, ΑΜΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν.
And since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is also divided into two triangles, and the angles KAM, KBM are right, therefore the remaining angles AKB, AMB are equal to two right angles.
εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΕΗ, ΔΕΖ δυσὶν ὀρθαῖς ἴσαι· αἱ ἄρα ὑπὸ ΑΚΒ, ΑΜΒ ταῖς ὑπὸ ΔΕΗ, ΔΕΖ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΑΚΒ τῇ ὑπὸ ΔΕΗ ἐστιν ἴση· λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση.
But the angles DEG, DEF are also equal to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the remaining angle AMB is equal to the remaining angle DEF.
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἡ ὑπὸ ΛΝΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση· καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση.
Similarly then it will be proved that the angle LNB is also equal to the angle DFE; therefore the remaining angle MLN is also equal to the remaining angle EDF.
ἰσογώνιον ἄρα ἐστὶ τὸ ΛΜΝ τρίγωνον τῷ ΔΕΖ τριγώνῳ· καὶ περιγέγραπται περὶ τὸν ΑΒΓ κύκλον.
Therefore the triangle LMN is equiangular to the triangle DEF; and it has been circumscribed about the circle ABC.
περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγέγραπται· ὅπερ ἔδει ποιῆσαι.
Therefore about the given circle a triangle equiangular to the given triangle has been circumscribed; which was required to do.
§4.prop.4εἰς τὸ δοθὲν τρίγωνον κύκλον ἐγγράψαι.
Into the given triangle to inscribe a circle.
ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ· δεῖ δὴ εἰς τὸ ΑΒΓ τρίγωνον κύκλον ἐγγράψαι.
Let ABC be the given triangle; it is required then into the triangle ABC to inscribe a circle.
τετμήσθωσαν αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δίχα ταῖς ΒΔ, ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας κάθετοι αἱ ΔΕ, ΔΖ, ΔΗ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΔ γωνία τῇ ὑπὸ ΓΒΔ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΕΔ ὀρθῇ τῇ ὑπὸ ΒΖΔ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΕΒΔ, ΖΒΔ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν ΒΔ· καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν·
Let the angles ABC, ACB be bisected by the straight lines BD, CD, and let them meet one another at the point D, and from D let DE, DF, DG be drawn perpendicular to the straight lines AB, BC, CA. And since the angle ABD is equal to the angle CBD, and the right angle BED is also equal to the right angle BFD, there are two triangles EBD, FBD having two angles equal to two angles, and one side equal to one side, namely their common side BD subtending one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; therefore DE is equal to DF.
ἴση ἄρα ἡ ΔΕ τῇ ΔΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΗ τῇ ΔΖ ἐστιν ἴση.
For the same reason also DG is equal to DF.
αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΔΕ, ΔΖ, ΔΗ ἴσαι ἀλλήλαις εἰσίν· ὁ ἄρα κέντρῳ τῷ Δ καὶ διαστήματι ἑνὶ τῶν Ε, Ζ, Η κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Η σημείοις γωνίας.
Therefore the three straight lines DE, DF, DG are equal to one another; therefore the circle described with center D and distance one of the points E, F, G will pass also through the remaining points, and will touch the straight lines AB, BC, CA because the angles at the points E, F, G are right.
εἰ γὰρ τεμεῖ αὐτάς, ἔσται ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπʼ ἄκρας ἀγομένη ἐντὸς πίπτουσα τοῦ κύκλου· ὅπερ ἄτοπον ἐδείχθη·
For if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle; which was proved absurd.
οὐκ ἄρα ὁ κέντρῳ τῷ Δ διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Η γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας· ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς τὸ ΑΒΓ τρίγωνον.
Therefore the circle described with center D and distance one of the points E, F, G will not cut the straight lines AB, BC, CA; therefore it will touch them, and the circle will be inscribed in the triangle ABC.
ἐγγεγράφθω ὡς ὁ ΖΗΕ.
εἰς ἄρα τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλος ἐγγέγραπται ὁ ΕΖΗ· ὅπερ ἔδει ποιῆσαι.
Let it be inscribed as FGE. Therefore into the given triangle ABC the circle EGF has been inscribed; which was required to do.